Consider the following unbalanced equation: Ca_{3} (PO_{4})_{2}(s)+3H_{2}SO_{4}(aq) \Rightarrow 3CaSO_{4}(s)+2H_{3}PO_{4}(aq). What

Question

Consider the following unbalanced equation: Ca_{3} (PO_{4})_{2}(s)+3H_{2}SO_{4}(aq) \Rightarrow 3CaSO_{4}(s)+2H_{3}PO_{4}(aq). What

sunshine022uv 2022-01-09 Answered

Consider the following unbalanced equation:

C a_{3} {(P O_{4})}_{2} (s) + 3 H_{2} S O_{4} (a q) \Rightarrow 3 C a S O_{4} (s) + 2 H_{3} P O_{4} (a q)

. What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 kg of calcium phosphate with 1.0 kg of concentrated sulfuric acid

(98 % H_{2} S O_{4}

by mass)?

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Expert Answer

Cassandra Ramirez

Answered 2022-01-10 Author has 30 answers

Balance the equation:

C a_{3} {(P O_{4})}_{2 (s)} + 3_{H 2} S O_{4 (a q)} \Rightarrow 3 C a S O_{4 (s)} + 2_{H 3 P O 4 (a q)}

Solve for the masses of the reactants:

H_{2} S O_{4}

1.00 k g H_{2} S O_{4} \times 0.98 \times \frac{1 m o l H_{2} S O_{4}}{98.09 \times 10^{- 3} k g H_{2} S O_{4}} = 10 . m o l H_{2} S O_{4}

C a_{3} {(P O_{4})}_{2}

1.00 k g C a_{3} {(P O_{4})}_{2} \times \frac{1 m o l C a_{3} {(P O_{4})}_{2}}{310.18 \times 10^{- 3} k g C a_{3} {(P O_{4})}_{2}}

= 3.2 m o l C a_{3} {(P O_{4})}_{2} \times \frac{3 m o l H_{2} S O_{4}}{1 m o l C a_{3} {(P O_{4})}_{2}} = 9.6 m o l H_{2} S O_{4}

Since

9.6 m o l < 10 m o l

,

C a_{3} {(P O_{4})}_{2}

is the limiting reactant.
Step 3
Finally, solve for the masses of the products:

3.2 m o l C a_{3} {(P O_{4})}_{2} \times \frac{3 m o l C a S O_{4}}{1 m o l C a_{3} {(P O_{4})}_{2}} \times \frac{135.15 g C a S O_{4}}{1 m o l C a S O_{4}}

= 1.3 \times 10^{3} g C a S O_{4}

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zurilomk4

Answered 2022-01-11 Author has 35 answers

First, balance the reaction.
$C a_{3} {(P O_{4})}_{2} (s) + 3 H_{2} S O_{4} (a q) \Rightarrow 3 C a S O_{4} (s) + 2 H_{3} P O_{4} (a q)$
The molar mass of calcium phosphate is $40.08 \times 3 + (30.97 + 16 \times 4) \times 2 = 310.18 \frac{g}{m} o l$
Therefore 1.0kg of calcium phosphate $= 1000 g$ , and dividing by 310.18g/mol gives 3.22 moles.
The molar mass of the sulfuric acid is $1.01 \times 2 + 32.07 + 16 \times 4 = 98.09 \frac{g}{m} o l$
The 1.0kg mixture is 98% sulfuric acid, which is 0.98kg or 980g. Dividing by the 98.09g/mol gives 9.99 moles.
From the balanced reaction, 1 mol calcium phosphate reacts with 3 mol sulfuric acid.
Therefore, 3.22 mol calcium phosphate will react with 3(3.22)=9.66mol sulfuric acid.
There are 9.99 mol sulfuric acid, so it is the excess reactant. We use the limiting reactant of calcium phosphate for the calculations.
From the reaction, 1 mol calcium phosphate forms 3 mol calcium sulfate and 2 mol phosphoric acid.
$(3.22 m o l C a_{3} {(P O_{4})}_{2}) (3) = 9.66 m o l C a S O_{4}$
$(3.22 m o l C a_{3} {(P O_{4})}_{2}) (2) = 6.44 m o l H_{3} P O_{4}$

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